Problem: Captain Brandon has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Michael and his merciless band of thieves. The Captain has probability $\dfrac{3}{7}$ of hitting the pirate ship. The pirate only has one good eye, so he hits the Captain's ship with probability $\dfrac{2}{7}$. If both fire their cannons at the same time, what is the probability that the pirate hits the Captain's ship, but the Captain misses?
If the Captain hits the pirate ship, it won't affect whether he's also hit by the pirate's cannons (and vice-versa), because they both fired at the same time. So, these events are independent. Since they are independent, in order to get the probability that the pirate hits the Captain's ship, but the Captain misses, we just need to multiply together the probability that the captain misses and the probability that the pirate hits. The probability that the Captian misses is $1 - $ (the probability the Captain hits), which is $1 - \dfrac{3}{7} = \dfrac{4}{7}$ The probability that the pirate hits is $\dfrac{2}{7}$ So, the probability that the pirate hits the Captain's ship, but the Captain misses is $\dfrac{4}{7} \cdot \dfrac{2}{7} = \dfrac{8}{49}$.